an observer stands 200 meters from the launch site of a hot-air balloon. balloon rises 4m/sec at a constant rate. how fast is the angle of elevation of the balloon increasing 30sec after the launch?

Question

anonymous · Answer

after 30sec, the balloon reaches a height of 120m.
the angle is then:
$$\alpha=\tan^{-1} (120/200) = 31degrees$$

anonymous · Answer

1.36

anonymous · Answer

sorry about the 1.36- I had a bad connection and couldnt finish.   $$d \theta/dt=4m/(200\sec ^{2}\theta)$$   and  $$\sec ^{2}\theta=1.36$$